5

$a_{n}$ sequence defined as :

$a_{n+1}=\dfrac{na_{n}+1}{a_{n}}$ , $a_0=1$

Then evaluate :

$\lim_{n\to\infty}n(n-a_{n})$

My attempt :

Call $\lim_{n\to\infty}a_{n}=L$ then I will use stolze Cesaro limit theorem

$\lim_{n\to\infty}n(n-a_{n})=\lim_{n\to\infty}\frac{n+1-a_{n+1}-n+a_{n}}{\frac{1}{n+1}-\frac{1}{n}}$

From here how can I complete ?

Tomas Houbaze
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3 Answers3

1

$a_{n+1}=\frac{na_n+1}{a_n}=n+\frac{1}{a_n}$ so that $n+1\ge a_{n+1} \gt n$. From this $a_n\to \infty$, as $n \to \infty$ Let $L_n=n(n-a_n)=n(1-\frac{1}{a_{n-1}})$. Since $a_n\to \infty$, $L_n\approx n \to \infty$.

herb steinberg
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1

The proof of Herb Steinberg is right (I don't know how to cite) $a_{n+1}=n+\frac{1}{a_n}$. Now, how $a_n>0$ we have $a_{n+1}>n$. Then, it is enough to prove that the limit of $b_n=n$ "goes to $\infty$" (squeeze theorem). Therefore the serie "does not converge" (Cauchy Criterion)

hugo d
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0

Pls do look at my comment too.

I think the answer is 1.Let me extract some stuff from herb's answer.Here $L_n= n(n-an)$ Now $L_n = n^2 - n*an$.

Replacing $n *an$ by $(a(n+1)*an)-1$.

$L_n = n^2 -[ (a(n+1)*an)-1]$.

$L_n =n^2 - (a(n+1)*an) +1$.

As herb specified that when $n$ approaches infinity $a(n+1) ~ an$ And similarly $(n+1) ~=n$.

Similarly $n < an < (n+1)$.

Which mean that $a(n+1)*an ~= n^2$. Which leaves $L_n$ as $1$.

lifeistod
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mechanics
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  • I do think that the sequence converge because of the fact that an - a(n-1) < 1. – mechanics Jul 13 '19 at 17:57
  • I simply say that let zn = n + (1/z(n-1)). When z(n-1) >1,then 1/z(n-1) decrease as z(n-1) increase . So when n increase by 1 then z(n+1) = n+1 +(1/z(n)) which mean that z(n+1) - z(n) = 1 - [(1/z(n)) -(1/z(n-1))] which always yeilds a result less than one. – mechanics Jul 13 '19 at 18:11
  • The sequence definitely diverges to $\infty$. I found that $a(n)$ is approximately $ n-1$ by a Python program. – Varun Vejalla Jul 14 '19 at 04:13